Chapter-17

Waves-II

Chapter-17 Waves-II

Topics to be studied Speed of sound waves Relation between displacement and

pressure amplitude Interference of sound waves Sound intensity and sound level Beats The Doppler effect

Longitudinal Waves: Particles displacement parallel to wave direction-Sound Waves

Wavefronts: Surfaces over which the oscillations have the same value. For point source such surfaces are represented by cirucles

Rays : lines representing the direction of sound wave.

Rays are to wavefronts

Ch 17-2 Sound Waves

Speed of Sound: speed of mechanical Wave v= (elastic property/inertial property)

A sound wave passes through medium, it undergoes compression and expansion due to pressure variation, then elastic property is due to change in volume or bulk modulus B=-p/(V/V) then

Speed of sound v = B/ where is density

Ch 17-3 Speed of Sound

Particle displacement s(x,t)=sm cos(kx-t)

where sm is displacement amplitude

Pressure variation given by

p= pmsin(kx-t)

where pm is pressure amplitude

Ch 17-4 Traveling Sound Wave

Ch 17-4 Traveling Sound Wave

Sound waves undergo interference if phase difference between two waves from s1 and s2 have phase difference =kx-t; k= 2/=2- 1= kL2-t-kL1+ t =k(L2-L1)

=k(L2-L1) = (2/ )L

path difference L =L2-L1 is multiple of wavelength

Fully Constructive Interference for L =n (n=0,1,2,3,….) Fully Destructive Interference for

L =m/2 (m=1,3,5,7…)

Ch 17-5 Interference

Ch 17-6 Intensity and Sound Level

Intensity I of sound is average rate of energy transferred by the wave through or onto the surface. If P is power and A is surface area (A=4R2 for a sphere) then

I=P/A=P/4R2 I=(v2s2

m)/2 Displacement Amplitude sm I

Ch 17-6 Intensity and Sound Level

The Decibel ScaleLarge variation in sound displacement amplitude:Loudest amplitude:10-5 m; Faintest amplitude: 10-

11 mSound intensity variation expressed in logarithms.Instead of sound intensity I, sound level given in

decibels (dB) by: = (10dB) log (I/I0), where I0 is standard

reference intensity I0 =10-12 W/m2

The 2- 1= (10dB) log (I2/I1)

Pipes resonates if An open end is an antinodes

and A closed end is a node For pipe open at both end:L=/2, 2/2, 3/2,….. = mm/2 (m=1,2,3,4,…)

fm=v/m=mv/2L (m=1,2,3,4,…) For pipe close at one end:L=/4, 3/4, 5/4,….. = nn/4 ( n=1,3,5,7,…)

fn=v/n=nv/4L (n=1,3,5,7,…)

Ch 17-7 Sources of Musical Sound

Ch 17-9 The Doppler Effect

The Doppler Effect : Change in observed frequency f’ with respect to source frequency f due to motion of source (vS) or detector (vD) or both:

f’=f(vvD)/(vvS)

When the detector or source are moving towards each other, the sign of speed must results in an increase in observed frequency f’.

When the detector or source are moving away from each other, the sign of speed must result in a decrease in observed frequency f’.

Ch 17-9 The Doppler Effect

Det. Moving in opposite direction-Source Stationary

Distance traveled by wavefront in t sec is vt and Distance traveled by detector in t sec in opposite direction is -vDt

Distance traveled by wavefront with respect to detector= vt-(-vDt)= vt+vDt

Number of wavelength intercepted by Detector= (vt+vDt)/

Observed frequency f’= Number of wavelength intercepted /t

f’= (1/t)(vt+vDt)/=(v+vD)/= f(v+vD)/v For detector moving in same

direction f’= (1/t)(vt-vDt)/=(v-vD)/= f(v-vD)/v

Source. Moving Det. Stationary

Source move towards detector with speed vS. During time T, the wavefront move a distance vT while the source move a distance vST. At the end of T, second sound Wavefront is emitted. The physical seperation between the two wavefront is ’=vT- vST

The observed frequency f’= v/’= f’= v/(vT- vST)=(v/T)(1/(v-vs)) f’= fv/(v-vs) Source moving from detector f’= fv/(v+vs)

f’= fv/(vvs)

Ch 17-9 The Doppler Effect

Ch 17-9 The Doppler Effect

Suggested problems: Chapter 17

  The quiz questions will be very similar to the following text-book problems. Refer to the course website for the latest version of this

document. You are encouraged to seek the help of your instructor during his office hours.   6. A man strikes one end of a thin rod with a hammer. The speed of sound in the rod is 15 times the speed of sound in air. A woman, at the

other end with her ear close to the rod, hears the sound of the blow twice with a 0.12 s interval between; one sound comes through the rod and the other comes through the air alongside the rod. If the speed of sound in air is 343 m/s, what is the length of the rod?

Answer: 44 m   12. The pressure in a traveling sound wave is given by the equation Δp = (1.50 Pa) sin {π[(0.900 m1) x (315 s1) t]}. Find the (a) pressure amplitude, (b) frequency, (c) wavelength, and (d) speed of the

wave. Answer: a) 1.50 Pa; b) 158 Hz; c) 2.22 m; d) 350 m/s   16. Two sound waves, from two different sources with the same frequency, 540 Hz, travel in the same direction at 330 m/s. The sources are

in phase. What is the phase difference of the waves at a point that is 4.40 m from one source and 4.00 m from the other? Answer: 4.11 rad = 236 º   22. In Fig. 17-37, sound with a 40.0 cm wavelength travels rightward from a source and through a tube that consists of a straight portion

and a half-circle. Part of the sound wave travels through the half-circle and then rejoins the rest of the wave, which goes directly through the straight portion. This rejoining results in interference. What is the smallest radius r that results in an intensity minimum at the detector?

  Answer: 17.5 cm   28. Two sounds differ in sound level by 1.00 dB. What is the ratio of the greater intensity to the smaller intensity? Answer: 1.26   58. A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 29.9 m/s, the

speed of surface B is 65.8 m/s, and the speed of sound is 329 m/s. The source emits waves at frequency 1200 Hz as measured in the source frame. In the reflector frame, what are the (a) frequency and (b) wavelength of the arriving sound waves? In the source frame, what are the (c) frequency and (d) wavelength of the sound waves reflected back to the source?

Answer: (a) 1.58×103 Hz; (b) 0.208 m; (c) 2.16×103 Hz; (d) 0.152 m   80. A detector initially moves at constant velocity vD directly toward a stationary sound source and then (after passing it) directly from it.

The emitted frequency is f. During the approach the detected frequency is fapp and during the recession it is frec. If the frequencies are related by (fapp - frec)/f = 0.500, what is the ratio vD/v of the speed of the detector to the speed of sound?

Answer: 0.25